हिंदी

Lim X → 0 Sin ( 2 + X ) − Sin ( 2 − X ) X

Advertisements
Advertisements

प्रश्न

\[\lim_{x \to 0} \frac{\sin \left( 2 + x \right) - \sin \left( 2 - x \right)}{x}\]

Advertisements

उत्तर

\[\lim_{x \to 0} \left[ \frac{\sin \left( 2 + x \right) - \sin \left( 2 - x \right)}{x} \right]\] 

\[\Rightarrow \lim_{x \to 0} \left[ \frac{2\cos \left( \frac{2 + x + 2 - x}{2} \right) \sin\left( \frac{2 + x - 2 + x}{2} \right)}{x} \right]\]
\[ \Rightarrow \lim_{x \to 0} \left[ \frac{2\cos 2 \sin x}{x} \right]\]
\[ \Rightarrow 2 \cos 2 \left[ \because \lim_{x \to 0} \frac{\sin x}{x} = 1 \right]\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.7 [पृष्ठ ५०]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.7 | Q 26 | पृष्ठ ५०

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\lim_{x \to 0} \frac{2 x^2 + 3x + 4}{x^2 + 3x + 2}\] 


\[\lim_{x \to 2} \left( 3 - x \right)\] 


\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]


\[\lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}\] 


\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]


\[\lim_{x \to 3} \left( x^2 - 9 \right) \left[ \frac{1}{x + 3} + \frac{1}{x - 3} \right]\] 


\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\] 


\[\lim_{x \to a} \frac{\left( x + 2 \right)^{5/2} - \left( a + 2 \right)^{5/2}}{x - a}\] 


\[\lim_{x \to a} \frac{x^{5/7} - a^{5/7}}{x^{2/7} - a^{2/7}}\] 


\[\lim_{x \to \infty} \frac{\left( 3x - 1 \right) \left( 4x - 2 \right)}{\left( x + 8 \right) \left( x - 1 \right)}\] 


\[\lim_{x \to \infty} \frac{5 x^3 - 6}{\sqrt{9 + 4 x^6}}\]


\[\lim_{x \to \infty} \sqrt{x^2 + 7x - x}\] 


\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\] 


\[\lim_{n \to \infty} \left[ \frac{1^2 + 2^2 + . . . + n^2}{n^3} \right]\]


\[\lim_{n \to \infty} \left[ \frac{1 + 2 + 3 . . . . . . n - 1}{n^2} \right]\] 


\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1


Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\] 


\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\] 


\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\] 


\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\] 


\[\lim_{x \to 0} \frac{1 - \cos 2x}{3 \tan^2 x}\] 


\[\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\]


Evaluate the following limits: 

\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\] 

 


Evaluate the following limit: 

\[\lim_{x \to 0} \frac{\sin\left( \alpha + \beta \right)x + \sin\left( \alpha - \beta \right)x + \sin2\alpha x}{\cos^2 \beta x - \cos^2 \alpha x}\]


\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]


\[\lim_{x \to 1} \frac{1 + \cos \pi x}{\left( 1 - x \right)^2}\] 


\[\lim_{x \to 1} \frac{1 - \frac{1}{x}}{\sin \pi \left( x - 1 \right)}\]


\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\] 


\[\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\]


Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]

 

Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]


\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]


Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]


\[\lim_{x \to 0} \frac{\sin 2x}{x}\] 


\[\lim_{x \to  } \frac{1 - \cos 2x}{x} is\]


The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\] 


\[\lim_{x \to 1} \left[ x - 1 \right]\] where [.] is the greatest integer function, is equal to 


Evaluate the following limit:

`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`


Evaluate the following limit:

`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`


Evaluate the Following limit:

`lim_(x->7)[[(root[3][x] - root[3][7])(root[3][x] + root[3][7])] / (x - 7)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×